Simplify; express your answer in exponential form. Assume $t\neq 0, z\neq 0$. $\dfrac{{(t^{-3})^{-1}}}{{(t^{-5}z^{2})^{2}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${t^{-3}}$ to the exponent ${-1}$ . Now ${-3 \times -1 = 3}$ , so ${(t^{-3})^{-1} = t^{3}}$ In the denominator, we can use the distributive property of exponents. ${(t^{-5}z^{2})^{2} = (t^{-5})^{2}(z^{2})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(t^{-3})^{-1}}}{{(t^{-5}z^{2})^{2}}} = \dfrac{{t^{3}}}{{t^{-10}z^{4}}}$ Break up the equation by variable and simplify. $\dfrac{{t^{3}}}{{t^{-10}z^{4}}} = \dfrac{{t^{3}}}{{t^{-10}}} \cdot \dfrac{{1}}{{z^{4}}} = t^{{3} - {(-10)}} \cdot z^{- {4}} = t^{13}z^{-4}$.